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Question

Find the unit vector perpendicular to each of the vectors 6^i+2^j+3^k and 3^i2^k.

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Solution

Vectorperpendicularto6^i+2^j+3^kc=(6^i+2^j+3^k)×(3^i2^k)=∣ ∣ ∣^i^j^k623302∣ ∣ ∣=^i(4)^j(129)+^k(06)=4^i+21^j6^kImplies,^c=4^i+2^j6^k16+(21)2+36=4^i+2^j6^k493

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